Integral of Brownian motion

Question

Is there any formula for the following integral ?

$$\int_0^t W_t^n\; \mathrm{d}W_t $$

For $n=1$ the answer is known. What about $n=2,3,\ldots$?

Answer 1

Using Ito's lemma with $f(x):=x^{n+1}$, $$ W_t^{n+1}=(n+1)\int_0^t W_s^n\,dW_s+\frac{(n+1)n}{2}\int_0^t W_s^{n-1}\,ds. $$ Therefore, $$ \int_0^t W_s^n\,dW_s=\frac{W_t^{n+1}}{n+1}-\frac{n}{2}\int_0^t W_s^{n-1}\,ds. $$